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(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
<h3>Total capacitance of the circuit</h3>
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
<h3>C1 and C2 are in series </h3>

<h3>C1 and C2 are parallel to C3</h3>

<h3>C(123) is series to C5 and C6</h3>

<h3>C7 and C8 are in series</h3>

<h3>Total capaciatnce of the circuit</h3>
Ct + C(78) = 2 μF + 3 μF = 5 μF
<h3 /><h3>Total charge stored in the circuit</h3>
The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
<h3>Charge stored in 3μF capacitor</h3>
Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
Learn more about capacitance of capacitor here: brainly.com/question/13578522
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The potential energy of the object depends on
- the height of the object with respect to some reference points,
- the mass of the object,
- the gravitational field the object is in.
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Hope it helps ~
Answer:
Hope i could help!
Explanation:
so all but one light could be burned out, and the last one will still function.